in a straight line is given by s=t sqrd -8t where s is in the feet
and t is the time in seconds the object has been in motion . how long ( to the nearist tenth ) will it take to move 15 feet
2006-08-20 18:31:45 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
It's just s=t²-8t to solve for t when
15 = t²-8t or t²-8t-15=0 and solving by the quadratic formula
x = (8±√(64+60))/2 => t=9.6 or t=-1.6
From the way the question was worded, I'd guess that the positive time is the one of interest.
Doug
2006-08-20 18:44:22 · answer #1 · answered by doug_donaghue 7 · 0⤊ 0⤋
Looks like Doug and Forever got this one, but I'll try it by completing the square.
s = t^2 - 8t where s = 15
t^2 - 8t = 15
t^2 - 8t + 16 = 15 + 16 = 31
(t - 4)^2 = 31
t = 4 + sqrt(31) because the negative root is of no interest.
t = 9.57 = 9.6 seconds (to the nearest tenth)
Same answer as they got; just a different way of doing it ...
2006-08-21 01:59:49 · answer #2 · answered by bpiguy 7 · 0⤊ 0⤋
Approx 9.57 seconds
Given 15 = T^2 - 8T , or T^2 - 8T -15 =0, you have 2 roots and one possible for your question.
2006-08-21 01:46:31 · answer #3 · answered by ForeverBrainless 2 · 0⤊ 0⤋
I think it may have something to do with factoring:
s=15
15=tsq-8t
t sq-8t-15=0
(t-3) x (t-5)=0
t-3=0 and t=3
t-5=0 and t=5
s=15 ft. when t=5 sec and t=3 sec
Maybe????
2006-08-21 01:49:45 · answer #4 · answered by Tony T 4 · 0⤊ 0⤋
4.79 seconds
2006-08-21 01:40:54 · answer #5 · answered by matthew c 2 · 0⤊ 0⤋
Dude, please stop with the physics!
2006-08-21 01:36:32 · answer #6 · answered by Anonymous · 0⤊ 0⤋